Chapter 9.2, Problem 4QQ

a.

To determine

The first term

Answer to Problem 4QQ

  $a_1=8$

Explanation of Solution

Given information : First term of the sequence is $a_2=-4$

5^th term of the sequence is $a_5=\frac{1}{2}$

Formula used : $a_n=a_1{r}^{n-1}$

Calculation :

  $\begin{array}{l}{a}_2=-4\Rightarrow a_{1}r=-4\text{ where }{a}_1={\text{1}}^{st}\text{term of the sequence}\text{.}\\ a_5=-\frac{1}{2}\Rightarrow a_1{r}^4=-\frac{1}{2}\\ \text{These two equations give }r=-\frac{1}{2}\\ \text{we have }{a}_{1}r=4\Rightarrow a_1\times \frac{1}{-2}=-4\Rightarrow a_1=8\end{array}$

b.

To determine

To find: common ratio.

Answer to Problem 4QQ

  $r=-\frac{1}{2}$

Explanation of Solution

Given information :

First term of the sequence is $a_2=-4$

5^th term of the sequence is $a_5=\frac{1}{2}$

Formula used : $a_n=a_1{r}^{n-1}$

Calculation:

  $\begin{array}{l}{a}_2=-4\Rightarrow a_{1}r=-4\text{ where }{a}_1={\text{1}}^{st}\text{term of the sequence}\text{.}\\ a_5=-\frac{1}{2}\Rightarrow a_1{r}^4=-\frac{1}{2}\\ \text{These two equations give }r=-\frac{1}{2}\\ \frac{a_1{r}^4}{a_{1}r}=-\frac{1}{8}\Rightarrow r=-\frac{1}{2}\end{array}$.

c.

To determine

To find: An explicit rule for the n^th term.

Answer to Problem 4QQ

  $a_n=8{\left(-\frac{1}{2}\right)}^{n-1}$

Explanation of Solution

Given information :

First term of the sequence is $a_2=-4$

5^th term of the sequence is $a_5=\frac{1}{2}$

Formula used : $a_n=a_1{r}^{n-1}$

Calculation :

This sequence is as follows,

  $\begin{array}{l}{a}_1=8\\ a_2=8\left(-\frac{1}{2}\right)=8{\left(-\frac{1}{2}\right)}^{2-1}\\ a_3=8\left(-\frac{1}{2}\right)=8{\left(-\frac{1}{2}\right)}^{3-1}\\ \text{Hence, }{a}_n=8{\left(-\frac{1}{2}\right)}^{n-1}\end{array}$

d.

To determine

To find: An recursive rule for the n^th term.

Answer to Problem 4QQ

  $a_n=a_{n-1}\left(r\right)$

Explanation of Solution

Given information : First term of the sequence is $a_2=-4$

5^th term of the sequence is $a_5=\frac{1}{2}$

Formula used : $a_n=a_1{r}^{n-1}$

  $\begin{array}{l}{a}_1=8\\ a_2=8\left(-\frac{1}{2}\right)=a_{1}r\\ a_3=8{\left(-\frac{1}{2}\right)}^2=a_{2}r\\ a_4=8{\left(-\frac{1}{2}\right)}^4=a_{3}r\\ \text{Hence, }{a}_n=a_{n-1}\left(r\right)\end{array}$